I am trying to solve this integral problem, but really can't figure it out. Some one please help me----

Integral (2x^3-7x^2+7x)/(2x-5)

The ans is-- [IMG]file:///C:/Users/SOWMIT%7E1/AppData/Local/Temp/moz-screenshot-1.png[/IMG][IMG]file:///C:/Users/SOWMIT%7E1/AppData/Local/Temp/moz-screenshot.png[/IMG]

2. Do you know how to do long division of polynomials?

3. Yes, so do I do long division and divide them up into small parts?

4. if you divide it all by 2x-5 you get
x^2-x+6-(30/2x-5) and you should be able to integrate this.

5. Originally Posted by skboss
Yes, so do I do long division and divide them up into small parts?
Yes.

6. Originally Posted by skboss
Yes, so do I do long division and divide them up into small parts?
$\frac{2x^3-7x^2+7x}{2x-5}=x^2-x+1+\frac{5}{2x-5}$

7. Originally Posted by ione
$\frac{2x^3-7x^2+7x}{2x-5}=x^2-x+1+\frac{5}{2x-5}$
Nice, I made a mistake by rushing it.